Problem: $\log_{2}32 = {?}$
Answer: If $\log_{b}x=y$ , then $b^y=x$ First, try to write $32$ , the number we are taking the logarithm of, as a power of $2$ , the base of the logarithm. $32$ can be expressed as $2\times2\times2\times2\times2$ $32$ can be expressed as $2^5$ $2^5=32$, so $\log_{2}32=5$.